The product of the ages of someone's children Maria's children are all in school - and their ages are all whole numbers.  If the school only takes children from $5$ up to $18$ years and the product of the children's ages is $60,060$ - how many children does Maria have?
Based on rules for divisibility, prime factorization and factoring, I keep getting $5$ children as the answer with all children always having unique ages.  These are the $3$ solutions of ages I got:
$5,6,11,13,14$
$5,7,11,12,13$
$6,7,10,11,13$
Are there any other solutions and how do you mathematically justify each solution and how to arrive at them?
 A: The prime factorization of the product gives us $$60060=2^2\times3\times5\times7\times11\times13.$$
Suppose Maria has four children ages $5,7,11,$ and $13$.  Since $2^2=4<5$ and $3<5$, these factors have to be distributed into the other ages of children.  As there is no way to fully distribute these factors in such a way to write $60060$ as a product of only four factors that all fall between $5$ and $18$, we can conclude that Maria has $5$ children, as you found.
A: The prime factorization of $60060$ is $2^{2}\times3\times5\times7\times11\times13$. Two of the children must be aged 13 and 11 respectively, because these factors cannot be combined with any other numbers to change them, or they will go outside of the 5-18 range. Since without the 11 and 13, there are 5 numbers left, there are somewhere between 3 and 7 children.
There cannot be exactly 7 children because then there will be 3 children below the age of 5, there cannot be 6 children either because in order to have 6 children, we have to combine two of the children that were underage in the 7 case, but that still leaves 1 underage child. This leaves the range to be 3-5 children
There cannot be 3 children because the factors other than 11 and 13 must form the third age, and the only possibility is 11, 13 and 420.
There cannot be 4 children because that will require the 5 factors other than 11 and 13 to form 2 numbers. Any three numbers from the remaining 5 will multiple to exceed the 18 limit, with the singular exception of $2\times2\times3$, but that case doesn't work because the other number would be 35.
